The integers $r$ and $k$ are randomly selected, where $-3 < r < 6$ and $1 < k < 8$. what is the probability that the division $r \div k$ is an integer value? express your answer as a common fraction.

Denote by the random variables representing the integer values , respectively. Then and , where denotes the discrete uniform distribution over the interval . So and have probability mass functions

We want to find , where is any integer.

We have six possible choices for :

(i) if , then is an integer when ;

(ii) if , then is an integer when ;

(iii) if , then is an integer when ;

(iv) if , then is an integer when ;

(v) if or , then is an integer only when in both cases.

If the selection of are made independently, then the joint distribution is the product of the marginal distribution, i.e.

That is, there are 48 possible events in the sample space. We counted 12 possible outcomes in which is an integer, so the probability of this happening is .

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

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step-by-step explanation: